Hi, can someone give me some help with the following?(adsbygoogle = window.adsbygoogle || []).push({});

The cosine Fourier series of period 2 for the function f(t) that takes the form f(t) = cosh(t-1) in the range [itex]0 \le t \le 1[/itex] is

[tex]

\cosh \left( {t - 1} \right) = \sinh \left( 1 \right)\left[ {1 + 2\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {n\pi t} \right)}}{{n^2 \pi ^2 + 1}}} } \right]

[/tex]

Setting t = 0 yields

[tex]\sum\limits_{n = 1}^\infty {\frac{1}{{n^2 \pi ^2 + 1}}} = \frac{1}{{e^2 - 1}}[/tex]...(1)

I need to deduce the values of the sum [tex]\sum\limits_{}^{} {\left( {n^2 \pi ^2 + 1} \right)^{ - 1} } [/tex] over odd n and even n.

I tried setting t = 1 so that I could evaluate the sum

[tex]

\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^n }}{{n^2 \pi ^2 + 1}}}

[/tex]...(2)

Once I did that, I added (2) to (1) and it looked like I would end up with just twice the sum of the even terms since the odd terms cancel. But that's not right since I don't get the correct answer when I divide the result by 2. Can someone suggest a way to do calculate the sum over odd n and over even n? Thanks.

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# Homework Help: Cosine Fourier seires of cosh(t-1)

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